Chapter 5, Problem 32P

To determine

The minimum coefficient of static friction needed if the car is to navigate the curve without slipping.

Answer to Problem 32P

The minimum coefficient of static friction needed if the car is to navigate the curve without slipping is $0.204$

Explanation of Solution

The figure 1 shows the free body diagram of the car.

Write an expression to calculate the net horizontal force.

  $\begin{array}{l}N\sin \theta +f_s\cos \theta =m{a}_x\\ N\sin \theta +{\mu }_{s}N\cos \theta =m\left(\frac{v^2}{r}\right)\end{array}$        (I)

Here, $N$ is the normal force, $\theta$ is the angle, $f_s$ is the static frictional force, ${\mu }_s$ is the coefficient of static friction, $m$ the mass of the car, $a_x$ is the horizontal acceleration, $v$ is the speed, and $r$ is the radius of path.

Write an expression for the net vertical force.

  $\begin{array}{l}N\cos \theta -mg-f_s\sin \theta =0\\ N\cos \theta -mg-{\mu }_{s}N\sin \theta =0\end{array}$        (II)

Here, $N$ is the normal force, $\theta$ is the angle, $f_s$ is the static frictional force, ${\mu }_s$ is the coefficient of static friction, $m$ the mass of the car, and $g$ is the acceleration due to gravity.

Neglect the frictional force and re-write equation (I)

  $\begin{array}{c}N\sin \theta +0\cos \theta =m\left(\frac{v^2}{r}\right)\\ N\sin \theta =m\left(\frac{v^2}{r}\right)\end{array}$        (III)

Neglect the frictional force and re-write equation (II)

  $\begin{array}{c}N\cos \theta -mg-0\sin \theta =0\\ N\cos \theta -mg=0\end{array}$        (IV)

Re-arrange equation (IV) to find $N$

  $N=\frac{mg}{\cos \theta }$        (V)

Substitute equation (V) in equation (III),

  $\begin{array}{c}\frac{mg}{\cos \theta }\left(\sin \theta \right)=m\left(\frac{v^2}{r}\right)\\ g\tan \theta =\left(\frac{v^2}{r}\right)\end{array}$        (VI)

Rewrite equation (VI) to find $\theta$

  $\theta ={\tan }^{-1}\left(\frac{v^2}{gr}\right)$        (VII)

Now consider the presence of frictional force.

Re-arrange equation (I) to find $\text{N}$

  $N=\frac{m{v}^2}{R\left(\sin \theta +{\mu }_s\cos \theta \right)}$        (VIII)

Re-arrange equation (II) to find $\text{N}$

  $N=\frac{mg}{\left(\cos \theta -{\mu }_s\sin \theta \right)}$        (IX)

Compare the expression (VIII) and (IX) and rearrange the equation to find ${\mu }_s$

  $\begin{array}{c}\frac{mg}{\left(\cos \theta -{\mu }_s\sin \theta \right)}=\frac{m{v}^2}{R\left(\sin \theta +{\mu }_s\cos \theta \right)}\\ {\mu }_s=\frac{v^2\cos \theta -gr\sin \theta }{v^2\left(\sin \theta +gr\cos \theta \right)}\end{array}$        (X)

Conclusion:

Substitute $15.0\text{m/s}$ for $v$, $9.80{\text{m/s}}^{\text{2}}$ for $g$ , $\text{75}\text{.0m}$  for $r$ in the equation (VII) to find $\theta$ 

   $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{{\left(15.0\text{m/s}\right)}^2}{\left(9.80{\text{m/s}}^{\text{2}}\right)\left(\text{75}\text{.0m}\right)}\right)\\ =17.0°\end{array}$

Substitute $20.0\text{m/s}$ for $v$, $9.80{\text{m/s}}^{\text{2}}$ for $g$ , $\text{75}\text{.0m}$  for $r$, and $17.0°$ for $\theta$  in the equation (XI) to find ${\mu }_s$ 

  $\begin{array}{c}{\mu }_s=\frac{{\left(20.0\text{m/s}\right)}^2\cos \left(17.0°\right)-\left(9.80{\text{m/s}}^{\text{2}}\right)\left(\text{75}\text{.0m}\right)\sin \left(17.0°\right)}{{\left(20.0\text{m/s}\right)}^2\left(\sin \left(17.0°\right)+\left(9.80{\text{m/s}}^{\text{2}}\right)\left(\text{75}\text{.0m}\right)\cos \left(17.0°\right)\right)}\\ =\frac{\text{382}{\text{.5m}}^{\text{2}}{\text{/s}}^{\text{2}}\text{-214}{\text{.9m}}^{\text{2}}{\text{/s}}^{\text{2}}}{\text{116}{\text{.9m}}^{\text{2}}{\text{/s}}^{\text{2}}\text{-702}{\text{.9m}}^{\text{2}}{\text{/s}}^{\text{2}}}\\ =0.204\end{array}$

 Therefore, the minimum coefficient of static friction needed if the car is to navigate the curve without slipping is $0.204$